MHB Proving $\sin\left(10^{\circ}\right)$ is Rational or Irrational

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The discussion centers on proving whether $\sin(10^{\circ})$ is rational or irrational. It utilizes the relation $\sin \alpha = 3\ \sin \frac{\alpha}{3} - 4\ \sin^{3} \frac{\alpha}{3}$, leading to the cubic equation $8\ x^{3} - 6\ x + 1 =0$. The argument posits that if $x$ were rational, it could be expressed as a fraction of integers, but this leads to a contradiction involving the square root of a non-integer. Consequently, the conclusion drawn is that $\sin(10^{\circ})$ is irrational. The discussion is supported by mathematical reasoning and a shared solution approach.
Fallen Angel
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Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it.
 
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Fallen Angel said:
Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it.

[sp]From the relation...

$\displaystyle \sin \alpha = 3\ \sin \frac{\alpha}{3} - 4\ \sin^{3} \frac{\alpha}{3}\ (1)$

... it follows that $\displaystyle x= \sin \frac{\pi}{18}$ must be root of the cubic equation...

$\displaystyle 8\ x^{3} - 6\ x + 1 =0\ (2)$

If x is rational, then You can write $\displaystyle x = \frac{a}{b}$, being a and b integers... but in this case for (2) it must be...

$\displaystyle b = 4\ \sqrt{3\ a - \frac{1}{2}}\ (3)$

... and that's impossible for a and b integers... the conclusion is that x isn't rational...[/sp]

Kind regards

$\chi$ $\sigma$
 
Good work chisigma, my solution was almost the same.

From $8x^3-6x+1=0$, let $y=2 sin \ 10º$, then $y^3-3y+1=0$ and this polynomial has no rational roots (because of Gauss Lemma).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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